Solve x+4/5xx=300-39

$$x + \frac { 4 } { 5 } x x = 300 - 39$$

Solve for x

$$x = \frac{\sqrt{20905} - 5}{8} \approx 17.448201847$$

$$x=\frac{-\sqrt{20905}-5}{8}\approx -18.698201847$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$x+\frac{4}{5}x^{2}=300-39$$

Subtract 39 from 300 to get 261.

$$x+\frac{4}{5}x^{2}=261$$

Subtract 261 from both sides.

$$x+\frac{4}{5}x^{2}-261=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{4}{5}x^{2}+x-261=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{5} for a, 1 for b, and -261 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{4}{5}\left(-261\right)}}{2\times \frac{4}{5}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{4}{5}\left(-261\right)}}{2\times \frac{4}{5}}$$

Multiply -4 times \frac{4}{5}.

$$x=\frac{-1±\sqrt{1-\frac{16}{5}\left(-261\right)}}{2\times \frac{4}{5}}$$

Multiply -\frac{16}{5} times -261.

$$x=\frac{-1±\sqrt{1+\frac{4176}{5}}}{2\times \frac{4}{5}}$$

Add 1 to \frac{4176}{5}.

$$x=\frac{-1±\sqrt{\frac{4181}{5}}}{2\times \frac{4}{5}}$$

Take the square root of \frac{4181}{5}.

$$x=\frac{-1±\frac{\sqrt{20905}}{5}}{2\times \frac{4}{5}}$$

Multiply 2 times \frac{4}{5}.

$$x=\frac{-1±\frac{\sqrt{20905}}{5}}{\frac{8}{5}}$$

Now solve the equation x=\frac{-1±\frac{\sqrt{20905}}{5}}{\frac{8}{5}} when ± is plus. Add -1 to \frac{\sqrt{20905}}{5}.

$$x=\frac{\frac{\sqrt{20905}}{5}-1}{\frac{8}{5}}$$

Divide -1+\frac{\sqrt{20905}}{5} by \frac{8}{5} by multiplying -1+\frac{\sqrt{20905}}{5} by the reciprocal of \frac{8}{5}.

$$x=\frac{\sqrt{20905}-5}{8}$$

Now solve the equation x=\frac{-1±\frac{\sqrt{20905}}{5}}{\frac{8}{5}} when ± is minus. Subtract \frac{\sqrt{20905}}{5} from -1.

$$x=\frac{-\frac{\sqrt{20905}}{5}-1}{\frac{8}{5}}$$

Divide -1-\frac{\sqrt{20905}}{5} by \frac{8}{5} by multiplying -1-\frac{\sqrt{20905}}{5} by the reciprocal of \frac{8}{5}.

$$x=\frac{-\sqrt{20905}-5}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{20905}-5}{8} x=\frac{-\sqrt{20905}-5}{8}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$x+\frac{4}{5}x^{2}=300-39$$

Subtract 39 from 300 to get 261.

$$x+\frac{4}{5}x^{2}=261$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{4}{5}x^{2}+x=261$$

Divide both sides of the equation by \frac{4}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{4}{5}x^{2}+x}{\frac{4}{5}}=\frac{261}{\frac{4}{5}}$$

Dividing by \frac{4}{5} undoes the multiplication by \frac{4}{5}.

$$x^{2}+\frac{1}{\frac{4}{5}}x=\frac{261}{\frac{4}{5}}$$

Divide 1 by \frac{4}{5} by multiplying 1 by the reciprocal of \frac{4}{5}.

$$x^{2}+\frac{5}{4}x=\frac{261}{\frac{4}{5}}$$

Divide 261 by \frac{4}{5} by multiplying 261 by the reciprocal of \frac{4}{5}.

$$x^{2}+\frac{5}{4}x=\frac{1305}{4}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{1305}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1305}{4}+\frac{25}{64}$$

Add \frac{1305}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{20905}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{20905}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{20905}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{20905}}{8} x+\frac{5}{8}=-\frac{\sqrt{20905}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{\sqrt{20905}-5}{8} x=\frac{-\sqrt{20905}-5}{8}$$